Parabola – JEE Main MathematicsPractice Questions & PYQs

A parabola is defined by the set of points equidistant from its focus and its directrix, with the vertex located exactly at the midpoint between these two features along the axis of symmetry. Because the focus is at the origin $(0,0)$ and the directrix is the vertical line $x=2$, the axis of symmetry is the $x$-axis, which intersects the directrix at the point $(2,0)$. The vertex is found by calculating the midpoint between the focus $(0,0)$ and this intersection point $(2,0)$, which gives $(\frac{0+2}{2}, \frac{0+0}{2})$ and results in $(1,0)$.

A key geometric property of parabolas is that the point of intersection of any two perpendicular tangents must always lie on the directrix. Given the parabola $y^2=8x$, we compare this to the standard form $y^2=4ax$ to determine that $4a=8$, which results in $a=2$. Consequently, the directrix is defined by the equation $x=-2$. Because the required point must exist on both the given tangent $y=x+2$ and the directrix, we substitute $x=-2$ into the line equation to obtain $y=0$, confirming the point is $(-2,0)$.

A parabola defined by the quadratic form $y = Ax^2 + Bx + C$ possesses a vertex at the coordinate $(-\frac{B}{2A}, f(-\frac{B}{2A}))$. Identifying the coefficients for the family of parabolas $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2 a$ gives $A = \frac{a^3}{3}$ and $B = \frac{a^2}{2}$, which establishes the x-coordinate of the vertex as $x = -\frac{a^2/2}{2(a^3/3)} = -\frac{3}{4a}$. Substituting this x-coordinate back into the original parabola equation yields $y = \frac{a^3}{3}(-\frac{3}{4a})^2 + \frac{a^2}{2}(-\frac{3}{4a}) - 2a$, which simplifies to $y = -\frac{35a}{16}$.

The path traced by the vertex as $a$ varies is found by eliminating the parameter $a$ using the expression $a = -\frac{3}{4x}$ derived from the x-coordinate. Substituting this into the simplified expression for $y$ results in $y = -\frac{35}{16}(-\frac{3}{4x})$, which leads to $y = \frac{105}{64x}$. The relationship governing the locus is therefore $xy = \frac{105}{64}$.

The locus of the midpoint of the segment connecting a fixed point to a variable point on a curve is found by expressing the coordinates of the variable point in terms of the midpoint's coordinates and substituting them into the original curve's equation. For the parabola $y^2=8x$ and the point $P(1, 0)$, we represent a generic point $Q$ on the parabola as $(x_1, y_1)$, which satisfies $y_1^2 = 8x_1$. The midpoint $(x, y)$ of the segment $PQ$ is given by $x = \frac{x_1 + 1}{2}$ and $y = \frac{y_1}{2}$. Rearranging these to solve for the original coordinates yields $x_1 = 2x - 1$ and $y_1 = 2y$. Substituting these expressions into the parabola equation results in $(2y)^2 = 8(2x - 1)$, which simplifies to $4y^2 = 16x - 8$. Dividing this equation by four provides the resulting locus $y^2 - 4x + 2 = 0$.

To determine the relationship between the coefficients of the line, we first identify the intersection points of the parabolas $y^2 = 4ax$ and $x^2 = 4ay$ by solving the system of equations simultaneously. The intersection coordinates are found to be the origin $(0, 0)$ and the point $(4a, 4a)$. Because the line $2bx + 3cy + 4d = 0$ must pass through both points, we substitute $(0, 0)$ into the linear equation, which immediately yields $4d = 0$, confirming that $d = 0$. Substituting the second point $(4a, 4a)$ into the line equation results in $2b(4a) + 3c(4a) + 4d = 0$, and with $d = 0$, this simplifies to $8ab + 12ac = 0$. Since $a \neq 0$, we can divide by $4a$ to obtain $2b + 3c = 0$, which collectively with $d=0$ leads to the expression $d^2 + (2b + 3c)^2 = 0$.

The equation for a parabola is standardly written as $y^2 = 4bx$, where any point on the curve can be represented by the parametric coordinates $(bt^2, 2bt)$. The slope of the tangent at $t_1$ is $1/t_1$, which makes the slope of the normal line $-t_1$, resulting in the normal equation $y = -t_1x + 2bt_1 + bt_1^3$. Since this normal line intersects the parabola again at a second distinct point defined by $t_2$, we substitute the coordinates $(bt_2^2, 2bt_2)$ into the normal equation to obtain $2bt_2 = -t_1(bt_2^2) + 2bt_1 + bt_1^3$.

Dividing throughout by $b$ simplifies the expression to $2t_2 = -t_1t_2^2 + 2t_1 + t_1^3$, which can be rearranged to $2(t_2 - t_1) = -t_1(t_2^2 - t_1^2)$. Factoring the difference of squares on the right side yields $2(t_2 - t_1) = -t_1(t_2 - t_1)(t_2 + t_1)$. Since the normal meets the parabola at a different point, we know $t_2 \neq t_1$, allowing us to divide by $(t_2 - t_1)$ to reach the result $2 = -t_1(t_2 + t_1)$. Rearranging this for $t_2$ gives the final relationship:
$t_2 = -t_1 - \frac{2}{t_1}$

Any tangent to the parabola $y^2 = 8ax$ can be represented by the equation $y = mx + \frac{2a}{m}$, where $m$ is the slope of the line. For this line to also be tangent to the circle $x^2 + y^2 = 2a^2$, the perpendicular distance from the circle's center at the origin $(0, 0)$ to the line $mx - y + \frac{2a}{m} = 0$ must equal the radius of the circle, which is $\sqrt{2a^2} = a\sqrt{2}$. This condition is expressed as:
$\left| \frac{2a/m}{\sqrt{m^2+1}} \right| = a\sqrt{2}$

Squaring both sides and simplifying the expression leads to $4 = 2m^2(m^2+1)$, which simplifies further to the biquadratic equation $m^4 + m^2 - 2 = 0$. Factoring this into $(m^2+2)(m^2-1) = 0$ reveals that $m^2 = 1$ is the only valid solution since $m^2$ must be positive, providing slopes of $m = 1$ and $m = -1$. Substituting these slopes back into the tangent line equation results in the two lines $y = x + 2a$ and $y = -x - 2a$, which are combined as $y = \pm(x + 2a)$.

To determine the common tangents of the parabola $y^2 = 4\sqrt{5}x$ and the circle $2x^2+2y^2=5$, we begin by expressing the general tangent to the parabola. Comparing the parabola equation to the standard form $y^2=4ax$, we identify $a=\sqrt{5}$, which gives the tangent line as $y=mx+\frac{\sqrt{5}}{m}$. For this line to also be tangent to the circle $x^2+y^2=\frac{5}{2}$, its perpendicular distance from the center $(0,0)$ must equal the radius, which is $\sqrt{\frac{5}{2}}$.

Applying the distance formula $\frac{|\frac{\sqrt{5}}{m}|}{\sqrt{m^2+1}} = \sqrt{\frac{5}{2}}$ and squaring both sides results in $\frac{5}{m^2(m^2+1)} = \frac{5}{2}$. Simplifying this expression leads to $m^2(m^2+1)=2$, or $m^4+m^2-2=0$. Factoring the quadratic gives $(m^2-1)(m^2+2)=0$, yielding the solutions $m=\pm 1$. Substituting $m=1$ into the tangent line equation confirms that $y=x+\sqrt{5}$ is a common tangent, thereby validating the first statement.

Regarding the second statement, the algebraic condition $m^4-3m^2+2=0$ is also satisfied by the valid tangent slopes $m=\pm 1$, since substituting $m^2=1$ into this equation yields $1-3+2=0$. Although this condition is satisfied by the correct slopes, it is mathematically distinct from the derived condition $m^4+m^2-2=0$ which defines all tangent possibilities. Consequently, while both statements are true, the general algebraic condition provided in the second statement does not directly explain the specific geometric existence of the tangent line in the first statement.

The geometry of the parabola $y^2 = 12x$ implies a focus at $S(3, 0)$ and allows any point on the parabola to be represented by the parametric coordinates $(3t^2, 6t)$. Defining the endpoints of the chord as $P(3t^2, 6t)$ and $Q(12t^2, 12t)$ ensures the ordinates remain in a $1:2$ ratio. The length of the chord is given by the distance formula:
$\sqrt{(12t^2-3t^2)^2 + (12t-6t)^2} = 3\sqrt{13}$
Squaring both sides of the equation yields:
$81t^4 + 36t^2 = 117$
which simplifies to the quadratic form:
$9t^4 + 4t^2 - 13 = 0$
Factoring this expression as $(t^2-1)(9t^2+13) = 0$ reveals that $t=1$ is the relevant positive solution.

With $t=1$, the coordinates of the endpoints are $P(3, 6)$ and $Q(12, 12)$. The angle $\alpha$ subtended at the focus $S(3, 0)$ is determined by the vectors $\vec{SP} = (0, 6)$ and $\vec{SQ} = (9, 12)$. The dot product of these vectors is:
$\vec{SP} \cdot \vec{SQ} = 72$
and their magnitudes are $|\vec{SP}| = 6$ and $|\vec{SQ}| = 15$. The cosine of this angle is calculated as:
$\cos \alpha = \frac{72}{6 \times 15} = \frac{4}{5}$
Applying the identity $\sin^2 \alpha + \cos^2 \alpha = 1$, it follows that $\sin \alpha = 3/5$.

The midpoints $A(0,1), B(1,1),$ and $C(1,0)$ define a triangle with vertices at $(0,2), (0,0),$ and $(2,0)$, resulting in side lengths of $2, 2,$ and $2\sqrt{2}$. Using the property for the incentre $D = \frac{a_1V_1 + a_2V_2 + a_3V_3}{a_1+a_2+a_3}$, the coordinates of $D$ simplify to $(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}})$. Substituting these coordinates into the parabola equation $y^2 = 4ax$ yields the relation $(\frac{2}{2+\sqrt{2}})^2 = 4a(\frac{2}{2+\sqrt{2}})$, which simplifies to $4a = \frac{2}{2+\sqrt{2}}$. Rationalizing the denominator gives $a = \frac{2-\sqrt{2}}{4} = \frac{1}{2} - \frac{1}{4}\sqrt{2}$. Comparing this to $\alpha + \beta\sqrt{2}$ yields $\alpha = \frac{1}{2}$ and $\beta = -\frac{1}{4}$, making the value of $\frac{\alpha}{\beta^2} = \frac{1/2}{1/16} = 8$.

The intersection of the circle $x^2+y^2=3$ and the parabola $x^2=2y$ determines the point $P$ where the line $L$ originates. Substituting the parabola's relationship $x^2=2y$ into the circle equation yields $y^2+2y-3=0$, which factors to $(y+3)(y-1)=0$. Since the problem specifies the first quadrant, we take $y=1$ and thus $x=\sqrt{2}$, establishing $P$ as $(\sqrt{2}, 1)$. By substituting these coordinates into the line equation $\sqrt{2}x + y = \alpha$, we calculate $\alpha = \sqrt{2}(\sqrt{2}) + 1 = 3$.

The tangency condition for a circle with radius $2\sqrt{3}$ and a center $Q(0, k)$ on the y-axis requires the perpendicular distance from the center to the line $\sqrt{2}x + y - 3 = 0$ to equal the radius. Using the distance formula, we solve:
$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$
This simplifies to $|k - 3| = 6$, yielding $k=9$ or $k=-3$, which places the centers at $Q_1(0, 9)$ and $Q_2(0, -3)$.

With vertices $P(\sqrt{2}, 1)$, $Q_1(0, 9)$, and $Q_2(0, -3)$, the triangle uses the segment $Q_1Q_2$ along the y-axis as its base, which has a length of $9 - (-3) = 12$. The height of the triangle from point $P$ to the y-axis is simply the x-coordinate of $P$, which is $\sqrt{2}$. Consequently, the area is $\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$, and the square of this area is $36 \times 2 = 72$.